Heat Chap02-068 - Chapter 2 Heat Conduction Equation 2-68 A...

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Chapter 2 Heat Conduction Equation 2-68 A compressed air pipe is subjected to uniform heat flux on the outer surface and convection on the inner surface. The mathematical formulation, the variation of temperature in the pipe, and the surface temperatures are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe. Properties The thermal conductivity is given to be k = 14 W/m °C. Analysis ( a ) Noting that the 85% of the 300 W generated by the strip heater is transferred to the pipe, the heat flux through the outer surface is determined to be 2 2 2 W/m 1 . 169 m) m)(6 (0.04 2 W 300 85 . 0 2 = × = = = π L r Q A Q q s s s Noting that heat transfer is one-dimensional in the radial r direction and heat flux is in the negative r direction, the mathematical formulation of this problem can be expressed as 0 = dr dT r dr d and - = - k dT r dr h T T r ( ) [ ( )] 1 1 k dT r dr q s ( ) 2 = ( b ) Integrating the differential equation once with respect to r gives r dT dr C = 1 Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, dT dr C r = 1 T r C r C ( ) ln = + 1 2 where C 1 and C 2 are arbitrary constants. Applying the boundary conditions give r = r 2 : k C r q C q r k s s 1 2 1 2 = = r = r 1 : k r q hr k r T C hr k r T C C r C T h r C k s 2 1 1 1 1 1 2 2 1 1 1 1 ln = ln )] ln ( [ - - - - = + - = - Substituting C C 1 2 and into the general solution, the variation of temperature is determined to be + + - = ° ° ° + + ° - = + + = + - + = - - + = 61 . 12 ln 483 . 0 10 C W/m 14 m) 04 . 0 )( W/m 1 . 169 ( m) C)(0.037 W/m 30 ( C W/m 14 ln C 10 ln ln ln ln ln ) ( 1 2 2 1 2 1 1 1 1 1 1 1 1 1 r r r r k r q hr k r r T C hr k r r T C hr k r T r C r T s 2-34 Heater L =6 m Air, -10°C r r 1 r 2
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Chapter 2 Heat Conduction Equation ( c ) The inner and outer surface temperatures are determined by direct substitution to be Inner surface ( r = r 1 ): ( 29 C 3.91 ° - = + + - = + + - = 61 . 12 0 483 . 0 10 61 . 12 ln 483 . 0 10 ) ( 1 1 1 r r r T Outer surface ( r = r 2 ): C 3.87 ° - = + + - = + + - = 61 . 12 037 . 0 04 . 0
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This note was uploaded on 08/24/2011 for the course ENGR 3150 taught by Professor Engel during the Spring '11 term at Georgia Southern University .

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Heat Chap02-068 - Chapter 2 Heat Conduction Equation 2-68 A...

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